Giải thích các bước giải:

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
{n_{{H_2}}} = 0,12mol\\
 \to {n_{Zn}} = {n_{{H_2}}} = 0,12mol\\
 \to {m_{Zn}} = 7,8g\\
 \to {m_{ZnO}} = 6,48g \to {n_{ZnO}} = 0,08mol\\
a)\\
\% {m_{Zn}} = \dfrac{{7,8}}{{14,28}} \times 100\%  = 54,62\% \\
\% {m_{ZnO}} = 100\%  - 54,62\%  = 45,38\% \\
b)\\
{n_{HCl}} = 2{n_{Zn}} + 2{n_{ZnO}} = 0,4mol\\
 \to {m_{HCl}} = 14,6g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{14,6 \times 100\% }}{{7,3\% }} = 200g\\
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} + {n_{ZnO}} = 0,2mol\\
 \to {m_{ZnC{l_2}}} = 27,2g\\
{m_{{\rm{dd}}X}} = {m_{hỗnhợp}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 214,04g\\
 \to C{\% _{ZnC{l_2}}} = \dfrac{{27,2}}{{214,04}} \times 100\%  = 12,7\% 
\end{array}\)